Advanced book on Mathematics Olympiad

526 Real Analysis

completing the induction. Hence the conclusion.

(USA Mathematical Olympiad, 2000, proposed by B. Poonen)

431.The casex=y=zis straightforward, so let us assume that not all three numbers

are equal. Without loss of generality, we may assume thatx≤y≤z. Let us first discuss

the casey≤x+y 3 +z. Theny≤x+ 2 z, and so

x+y+z

3

≤

x+z

2

≤z.

Obviouslyx≤(x+y+z)/3, and consequently

x+y+z

3

≤

y+z

2

≤z.

It follows that there exists, t∈[ 0 , 1 ]such that

x+z

2

=s

x+y+z

3

+( 1 −s)z,

y+z

2

=t

x+y+z

3

+( 1 −t)z.

Adding up these inequalities and rearranging yields

x+y− 2 z

2

=(s+t)

x+y− 2 z

3

.

Sincex+y< 2 z, this equality can hold only ifs+t=^32. Writing the fact thatfis a

convex function, we obtain

f

(

x+z

2

)

=f

(

s

x+y+z

3

+( 1 −s)z

)

≤sf

(

x+y+z

3

)

+( 1 −s)f (z),

f

(

y+z

2

)

=f

(

t

x+y+z

3

+( 1 −t)z

)

≤tf

(

x+y+z

3

)

+( 1 −t)f(z),

f

(

x+y

2

)

≤

1

2

f(x)+

1

2

f(y).

Adding the three, we obtain

f

(

x+y

2

)

+f

(

y+z

2

)

+f

(

z+x

2

)

≤(s+t)f

(

x+y+z

3

)

+

1

2

f(x)+

1

2

f(y)+( 2 −s−t)f(z)

=

2

3

f

(

x+y+z

3

)

+

1

2

f(x)+

1

2

f(y)+

1

2

f(z),